Why can no one in sci.math understand my simple point?
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From: Sylvia Else on 22 Jun 2010 02:31 On 22/06/2010 10:44 AM, Sylvia Else wrote: > On 15/06/2010 2:13 PM, |-|ercules wrote: >> Consider the list of increasing lengths of finite prefixes of pi >> >> 3 >> 31 >> 314 >> 3141 >> .... >> >> Everyone agrees that: >> this list contains every digit of pi (1) >> >> as pi is an infinite digit sequence, this means >> >> this list contains every digit of an infinite digit sequence (2) >> >> similarly, as computable digit sequences contain increasing lengths of >> ALL possible finite prefixes >> >> the list of computable reals contain every digit of ALL possible >> infinite sequences (3) > > The discussion on permutations of lists of computable reals to construct > diagonals showed that it is possible to construct diagonals that have > any finite prefix. However, you were quick to point out that as the > computable reals contain numbers like 0.111111...., diagonals not > containing a 1 are impossible. > > That is to say, there is a set of reals which contains all possible > finite prefixes, but which does not contain all possible infinite > sequences. > > Since that contradicts the reasoning used to conclude your proposition > (3), you'd have to prove it some other way. > > Sylvia. Looks like this is going to be ignored because it's inconvenient. Sylvia.
From: WM on 22 Jun 2010 05:34 On 22 Jun., 01:29, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > > Correct. Therefore the set of antidiagonals appears to be uncountable > > = unlistable. But it is countable. > > I don't see why you say it appears to be uncountable. (See below) > > > The set of antidiagonals that can be constructed (by a given > > substitution rule) from an infinite sequence of lists. > > Ah, OK. > > So we have (L0, L1, L2, ...), and corresponding to each Ln we have an > antidiagonal An. So we have a sequence (A0, A1, A2, ...). > > But (A0, A1, A2, ...) is obviously countable. Above you say it's "certainly > not countable", but it is. The set is certainly countable. But it cannot be written as a list because the antidiagonal of the supposed list would belong to the set but not to the list. Therefore it is not countable. Regards, WM
From: WM on 22 Jun 2010 05:38 On 22 Jun., 01:32, Virgil <Vir...(a)home.esc> wrote: > In article > <e0b8cbfa-70a9-4be3-a08f-117e0af7f...(a)q12g2000yqj.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 21 Jun., 22:17, Virgil <Vir...(a)home.esc> wrote: > > > > > But the set of all diagonals constructed according to the scheme given > > > > above is countable but cannot be listed. > > > > Whatever do yu mean by "the set of all diagonals"? > > > If you mean one "diagonal" for each possible list, that set of diagonals > > > is not countable. > > > If you mean the set of all "diagonals" for a single list, that is not > > > countable either, since each of uncountably many permutations of the > > > list produces a different "diagonal". > > > Take a list of all rationals. Construct, accordingto a given > > substitution rule the antidiagonal. Add it at first position to the > > list. Construct, according to the given rule the antidiagonal, add > > it ... and so on. > > You are saying given a list, create its anti-diagonal and prepend it to > that list. Repeat with the new list ad infinitum. > > > > > The number of antidiagonal is countable. Nevertheless it cannot be > > listed. > > I see no problem in listing the set of such anti-diagonals. Including all rational numbers! > They can > even be listed in the order in which each anti-diagonal is to be created. > > Note that the list of anti-diagonals so far created is finite so long as > the number of iterations is finite, and only becomes infinite when the > process achieves infinitely many iterations, and thus it is countable. > > Which such anti-diagonals does WM claim remain unlisted? That of the intended list of all those antidiagonals and all rational numbers. > > > The complete infinite binary tree contains, by definition, all real > > numbers between 0 and 1 as infinite paths, i. e., as infinite > > sequences { 0, 1 }^N of bits. > > > 0, > > / \ > > 0 1 > > / \ / \ > > 0 1 0 1 > > / > > 0 ... > > > The set { K_k | k in N } of nodes K_k of the tree is countable. > > > K_0 > > / \ > > K_1 K_2 > > / \ / \ > > K_3 K_4 K_5 K_6 > > / > > K_7 ... > > > The tree is constructed by extending the configurations B_j as > > explained below: > > > _________________ > > B_0 = > > > K_0 > > _________________ > > B_1 = > > > K_0 > > / > > K_1 > > _________________ > > B_2 = > > > K_0 > > / \ > > K_1 K_2 > > _________________ > > B_3 = > > > K_0 > > / \ > > K_1 K_2 > > / > > K_3 > > _________________ > > > B_4 = > > > K_0 > > / \ > > K_1 K_2 > > / \ > > K_3 K_4 > > _________________ > > ... > > _________________ > > B_j = > > > K_0 > > / \ > > K_1 K_2 > > / \ > > K_3 K_4 ... > > ... > > ... K_j > > _________________ > > ... > > _________________ > > > The complete infinite binary tree (including all those infinite paths > > which consist of nodes and edges only) is constructed by a countable > > number of steps. In no step more than one infinite paths is extended. > > Hence there are not more than countably many infinite paths. > > I have no idea what sort of definition of "countability" of infinite > sets that WM is using, I know that you cannot read the above text. But perhaps somewhat else can. Regards, WM
From: Sylvia Else on 22 Jun 2010 05:49 On 22/06/2010 7:34 PM, WM wrote: > On 22 Jun., 01:29, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: >> "WM"<mueck...(a)rz.fh-augsburg.de> wrote in message >> > >>> Correct. Therefore the set of antidiagonals appears to be uncountable >>> = unlistable. But it is countable. >> >> I don't see why you say it appears to be uncountable. (See below) >> > >>> The set of antidiagonals that can be constructed (by a given >>> substitution rule) from an infinite sequence of lists. >> >> Ah, OK. >> >> So we have (L0, L1, L2, ...), and corresponding to each Ln we have an >> antidiagonal An. So we have a sequence (A0, A1, A2, ...). >> >> But (A0, A1, A2, ...) is obviously countable. Above you say it's "certainly >> not countable", but it is. > > The set is certainly countable. But it cannot be written as a list > because the antidiagonal of the supposed list would belong to the set > but not to the list. Therefore it is not countable. > > Regards, WM The antidiagonal of the list of (A0, A1, A2,...) would only belong to the set if it is also the antidiagnoal of some Ln, which you haven't proved to be the case. Sylvia.
From: WM on 22 Jun 2010 07:18
On 22 Jun., 11:49, Sylvia Else <syl...(a)not.here.invalid> wrote: > >> So we have (L0, L1, L2, ...), and corresponding to each Ln we have an > >> antidiagonal An. So we have a sequence (A0, A1, A2, ...). > > >> But (A0, A1, A2, ...) is obviously countable. Above you say it's "certainly > >> not countable", but it is. > > > The set is certainly countable. But it cannot be written as a list > > because the antidiagonal of the supposed list would belong to the set > > but not to the list. Therefore it is not countable. > > The antidiagonal of the list of (A0, A1, A2,...) would only belong to > the set if it is also the antidiagnoal of some Ln, which you haven't > proved to be the case. There is a sequence of lists including a sequence of diagonals. Every list Ln includes the diagonals A0 to A(n-1). There is no limit. Not every sequence has a limit, in particular the sequence 1, 2, 3, ... does not have a limit. But as the list is and remains countably infinite, there is no problem if you would like to have a limit. There is none, but if there was one, the list would maintain its same structure. Regards, WM |